$\begin{aligned}
& \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \
& \mathbf{p O H}=4.745
\end{aligned}$
on adding 0.05 mole HCl
$\begin{aligned}
& \begin{array}{llll}
\mathrm{NH}_3+ & \mathrm{H}^{\oplus} & \rightarrow & \mathrm{NH}_4^{\oplus} \
0.1 & 0.05 & & 0.1 \
0.05 & 0 & & 0.15
\end{array} \
& \mathrm{pOH}^{\prime}=4.745+\log 3 \
& \mathrm{pOH}^{\prime}-\mathrm{pOH}=0.477 \
& 14-\mathrm{pH}^{\prime}-14+\mathrm{pH}=0.477 \
& \Delta \mathrm{pH}=0.477 \
& =47.7 \times 10^{-2} \approx 48 \times 10^{-2}
\end{aligned}$